# Talk:Ragged holes

You can't really use CEP(p) since you have no idea of what percentage of holes you'll be able to find if you shoot another target.

Consider this situation. You make 100 shots. There are enough shots for which a position can be determined so that you can estimate the group center. Now you count how many shots you can find (say 30). Take half of that number-15. Now use CEP(85). Sounds good right? Dah... How many different circles can you draw that eliminate 15 of the holes? A lot. It would conceivable to find all such circles and take average position and size. Not possible work for a human, but a computer doesn't care.

If done right ragged hole measurements aren't crippling.

Herb (talk) 18:12, 1 June 2015 (EDT)

*Herb: Since you took out the following you need to provide some sort of substantial criticism of it. I can't tell if what you say above is an attempt to explain it or disprove it? More than one person has agreed that this is the best practical solution to the problem it addresses:*

**In practice, given a target with a ragged hole and a small number of "censored" shots, it is probably adequate to place them evenly inside the hole. If the number of censored shots is large a better solution is to:****Set***p*= proportion of shots that were censored.**Find the smallest sigma such that CEP(p) covers the ragged hole.**

- I was arguing against that part in the comments above. You want a measurement that is repeatable. You could make the measurement as you suggested then use the experimental CEP(p) to estimate CEP(50). That would at least give you a consistent measure.

- To me this is a convex hull type problem. Two points will probably determine a circle, at most three. More on the circle just means that the position of the centers of the shots is within the measurement slop. So you really have a very few shots on the ragged hole which will determine where the center is. In part think of this as a sighting problem where center of the line segment for the extreme spread measurement controls "sighting" the weapon. Does that give you a warm fuzzy?

- It all is a matter of how many shots were taken and what percentage of the shots are in the ragged hole and so on. For example is there another hole on the target which is large and multiple shots could have snuck in? Think of it as something between covering circle in the worse case and a true CEP(50) in the best case. So even though you could consistently estimate CEP(50) you aren't consistently measuring CEP(50). So would either option work with 10 shots where 8 are in the ragged hole?

- Realistically no one is going to be shooting a 100 shots at a target and left wondering where the 46 that went in the ragged hole were. 5-10 shots is probably more the norm with maybe 1-3 "disappearing." Small sample statistics suck.

**Do you want it back in?**If so how about:

**In practice, given a target with a ragged hole it is probably adequate to take one of the two options.**

**(1) If there are a small number of "censored" shots, place them "evenly" inside the hole.****(2) If the number of censored shots is large then a better solution is to:****Find the smallest hole such that covers the ragged hole.****From***n*, the number of shots fired, and*c*the number outside the circle, determine CEP(n-c) for the number of holes censored by the circle.**Use the experimental CEP(p) measurement to estimate CEP(50) so that the measurement is repeatable.**

- To perhaps express it better, most shooters will be taking a small number of shots. Think of the average radial difference between the median shot and the next larger shot as the "granularity" of the CEP(50) measurement. Getting a "good" CEP number essentially relies on the fact that a "large" sample is used so that the "granularity" is small.

Herb (talk) 01:46, 2 June 2015 (EDT)

- To perhaps express it better, most shooters will be taking a small number of shots. Think of the average radial difference between the median shot and the next larger shot as the "granularity" of the CEP(50) measurement. Getting a "good" CEP number essentially relies on the fact that a "large" sample is used so that the "granularity" is small.

*I don't understand your variant of the proposal (#2 above). CEP(*x*) is supposed to denote a fraction*x*, so I don't know what CEP(n-c) is referring to. I also don't know what the experimental CEP(p) measure is referring to, or how one uses it to make a "repeatable CEP estimate."*

*I also don't understand the criticism: I do see 100-shot targets with ragged holes. Just because someone can't hang or change a lot of targets doesn't mean we can't say something strong about their precision when they have a sample with large*n*. The attraction of the original proposal is that we are imposing the same model on the censored shots as we use for observed shots. David (talk) 10:40, 2 June 2015 (EDT)*

- Fair enough. Change the last line in the text I proposed to
**From n, the number of shots fired, and c the number outside the circle, determine p=(n-c)/n that the radius of CEP(p) for the number of holes censored by the circle. Use the experimental CEP(p) measurement to estimate CEP(50) so that the measurement is repeatable.**

- Fair enough. Change the last line in the text I proposed to

- The whole point of converting to CEP(50) was just to have consistency. If you get CEP(93) on one target and CEP(87) on another and CEP(85) on another how do you compare the circle radii?

- The number of total shots is nebulous as the text is now written. So if the idea was to give a cookbook solution, it just doesn't seem to work to me.

- For the 100-shot target that you showed above I could see the position of about a dozen shots. Most of those would be within a circle that covered the ragged hole I think. So the measure of that target is much closer to a covering circle than a CEP(50) measurement - ie the precision of the measurement will be about that of the covering circle measurement. Obviously for a different pattern of 100 shots the situation would be different. With at least 20 of the 100 holes outside the circle you could get a pretty good CEP((50) estimate. If you could find 60 holes then you still couldn't get a true CEP(50) since you can't use 40 of the holes to help locate the center. But the 60 holes that you can find would give you a very reasonable estimate. You obviously could create a number of different cases to simulate.

**OK, yes, we should provide the math for this. CEP(***p*) is just another expression for the Rayleigh CDF. Given a*sample*CEP(*p*) I think we will need to apply a correction factor that's a function of*n*– probably \(c_{R}(n)\) – to back into an estimate of the Rayleigh parameter. Whoever gets this first please post it! David (talk) 14:03, 2 June 2015 (EDT)